Answer:
No. See the explanation below.
Explanation:
No. When we have the lineal model given by:
![Y_i = \beta_0 +\beta_1 X_i +\epsilon_i , i = 1,....,n](https://img.qammunity.org/2021/formulas/mathematics/college/90kdc17grzct29ujzc2k453o7u7j3njtwv.png)
For n observations, where y represent the dependent variable, X represent the independent variable and
are the parameters of the model, we are assuming that
is and independent and identically distrubuted variable that follows a normal distribution with the following parameters
.
So then the expected value for any error term is
![E(\epsilon_i) =0, i =1,...,n](https://img.qammunity.org/2021/formulas/mathematics/college/g46kicg8sxxhsbhwbb4z6f1ry3j73b7v85.png)
So then if we find the expected value for any observation we have this:
![E(Y_i) = E(\beta_0 +\beta_1 X_i +\epsilon_i) , i = 1,....,n](https://img.qammunity.org/2021/formulas/mathematics/college/lcfy917jxdkl9suadilvnevs39ak0dadfk.png)
Now we can distribute the expected value on the right by properties of the expected value like this:
![E(Y_i) = E(\beta_0) +E(\beta_1 X_i) +E(\epsilon_i), i =1,...,n](https://img.qammunity.org/2021/formulas/mathematics/college/l11ufhbtvnq5gp7da6cdppdy0foxcrd92f.png)
By properties of the expected value
if a is a constant and X a random variable, so then if we apply this property we got:
![E(Y_i) = \beta_0 +\beta_1 E(X_i) +0 ,i=1,...,n](https://img.qammunity.org/2021/formulas/mathematics/college/t172fge0maxbvyhsdoypeuslq9tmgjy3y5.png)
![E(Y_i) =\beta_0 +\beta_1 X_i, i=1,.....,n](https://img.qammunity.org/2021/formulas/mathematics/college/teld0dfi23tmizj778jbql8urc6myesi9l.png)
And if we see that'ts not the result supported by the claim for this reason is FALSE the statement.